Question

For an optical communication system, operating at $\lambda =800$nm, only $1\%$ of the optical source frequency is the available channel band width. How many channels can be accommodated for transmitting video T.V. signal requiring an approximate band width of $4.5$MHz?

• A. $8.3\times {10}^7$
• B. $8.3\times {10}^5$
• C. $1.71\times {10}^{20}$
• D. $1.71\times {10}^{22}$

Answer 1

Answer: (B)

$8.3\times {10}^5$

Optical source frequency,
$\nu =\frac{c}{\lambda }=\frac{3\times {10}^8}{800\times {10}^{-9}}=3.75\times {10}^{14}$Hz.
Band width of channel $=1\%$ of the source frequency
$=\frac{1}{100}\times 3.75\times {10}^{14}=3.75\times {10}^{12}$Hz
Number of channels for video T.V. signal
$=\frac{3.75\times {10}^{12}}{4.5\times {10}^6}=8.3\times {10}^5$.